Let $a = dx$ and $b = dy$ with $(x, y) = 1$ We need to prove $y = 1$ $ad + bdxy < \frac{5d^2xy}{2}$ $a < xy(\frac{5d}{2} - b)$ If $y \geq 3$, then $b \geq 3d$ $0 > \frac{5d}{2} - b > a$, a contradiction. So we only need to consider $y = 2$. Then $2x(\frac{5d}{2} - 2d) = 2x(\frac{d}{2}) = dx = a > a$, a contradiction. Hence $y = 1$ which means that $b \mid a$

Let $a = dx$ and $b = dy$ with $(x, y) = 1$ $d^2x+d^2xy^2< \frac{5d^2xy}{2}$ $y^2+1<\frac{5y}{2} \to y \in (\frac{1}{2},2) \to y=1$